Hq+k=y, for q helppp

8,4,6,2 witch is not an composite number

liberstina [14]

Composite numbers are numbers that have more than 2 factors. With that being said, your answer would be 2.

4 is a composite number: 1, 2, 4

6 is a composite number: 1, 2, 3, 6

8 is a composite number: 1, 2, 4, 8

2 is a prime number: 1 and 2

2 is a prime number, not a composite number.

7 0

2 years ago

Michael wants to find the length of the shadow of a tree. He measures the height of a fencepost and the length of the shadow it

zmey [24]

The shadow of the tree is about 17.955 feet long.

4 0

2 years ago

Carmen is writing an article for a magazine. She will be paid a fee and also will be paid for each word that is published. The t

enot [183]

Answer:

The inverse function of f(x)=2.5*x+150 is f⁻¹(x)= $\frac{2}{5}x-60$

Step-by-step explanation:

An inverse or reciprocal function of f (x) is called another function f ⁻¹(x) that fulfills that:

If f(a)=b then f⁻¹(b)=a

That is, inverse functions are functions that do the "opposite" of each other. For example, if the function f (x) converts a to b, then the inverse must convert b to a.

To construct or calculate the inverse function of any function, you must follow the steps below:

Since f (x) or y is a function that depends on x, the variable x is solved as a function of the variable y. And since inverse functions swap the input and output values (that is, if f (x) = y then f⁻¹(y) = x), then the variables are swapped and write the inverse as a function.

You know that he function f(x) = 2.5*x + 150 or y=2.5*x +150

Solving for x:

2.5*x +150=y

2.5*x= y-150

$x=\frac{y-150}{2.5}$

$x=\frac{y}{2.5}-\frac{150}{2.5}$

$x=0.4y-60=\frac{2}{5}y-60$

Exchanging the variable, you obtain that the inverse function of f(x)=2.5*x+150 is f⁻¹(x)= $\frac{2}{5}x-60$

7 0

3 years ago

For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?

Bond [772]

Given that $r=\theta$ , then $r'=1$

The slope of a tangent line in the polar coordinate is given by:

$m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$

Thus, we have:

$m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}$

Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

$\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

θ = 0

θ = 2.02875783811043

θ = 4.91318043943488

Part B:

For vertical tangent lines, $\frac{1}{m} =0$

Thus, we have:

$\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

θ = 4.91718592528713

3 0

3 years ago

Asa Sss Sss Sas Aas Hl Not congruent

ELEN [110]

Answer:

HL is the correct answer of this question

4 0

2 years ago